Regression of p nl/(n1+n2)on a Slope=-.09553,std. error=.0448,Z=2.132 Overall chi2(2) 4.551,pr>chi2=0.1027 Chi2 (1)for trend 4.546,pr>chi2=0.0330 Chi2(1) for departure 0.004, pr>chi2 =0. 9467 The"Chi2(1)for trend" is slightly different It's 4.546 rather than 4.515 Well, ptrend is just using N rather thann -1 in the formula Qtrend Chi2(1) for trend =N*r Let's go back to data arranged for the corr computation and show thi quietly corr y a [fw=weight] display r(N)*r(rho)2 4.5464579 slope=0 is given by the Qtrend statistic Well, we all know the relationship between Pearson,s correlation and regression-this is all this is Qdeparture("Chi2 (1)for departure"as Royston's output nicely labels it) is the statistic for the Cochran-Armitage test. But Trend and Dep arture are usually performed at the same time, so lumping them together under the name"Cochran-Armitage"is sometimes loosely done. The null hypothesis for the Cochran-Armitage test is that the trend is linear, and the test is for"departures"from linearity; i.e., it's simply a goodness -of-fit test for the linear model Qs(or equivalently Qtrend ) tests the null hypothesis of no associ ation. Since it's just a Pearson,s correlation we know that it's powerful ag ainst alternative hypotheses of monotonic trend but it's not at all powerful against curvilinear (or other) associations with a0 linear component. Model it Rich Goldstein recommended log istic regression Regression is certainly a better context to understand what you are doing -rather than all these chi-squared tests that are simply Pearson s correlations or goodness -of-fitRegression of p = n1/(n1+n2) on a: Slope = -.09553, std. error = .0448, Z = 2.132 Overall chi2(2) = 4.551, pr>chi2 = 0.1027 Chi2(1) for trend = 4.546, pr>chi2 = 0.0330 Chi2(1) for departure = 0.004, pr>chi2 = 0.9467 The “Chi2(1) for trend” is slightly different. It’s 4.546 rather than 4.515. Well, ptrend is just using N rather than N − 1 in the formula: Qtrend = Chi2(1) for trend = N * ray 2 Let’s go back to data arranged for the corr computation and show this. . quietly corr y a [fw=weight] . display r(N)*r(rho)^2 4.5464579 Qtrend is just Pearson’s correlation again. A regression is performed here to compute the slope, and the test of slope = 0 is given by the Qtrend statistic. Well, we all know the relationship between Pearson’s correlation and regression—this is all this is. Qdeparture (="Chi2(1) for departure" as Royston’s output nicely labels it) is the statistic for the Cochran–Armitage test. But Qtrend and Qdeparture are usually performed at the same time, so lumping them together under the name “Cochran–Armitage” is sometimes loosely done. The null hypothesis for the Cochran–Armitage test is that the trend is linear, and the test is for “departures” from linearity; i.e., it’s simply a goodness-of-fit test for the linear model. Qs (or equivalently Qtrend) tests the null hypothesis of no association. Since it’s just a Pearson’s correlation, we know that it’s powerful against alternative hypotheses of monotonic trend, but it’s not at all powerful against curvilinear (or other) associations with a 0 linear component. Model it Rich Goldstein recommended logistic regression. Regression is certainly a better context to understand what you are doing—rather than all these chi-squared tests that are simply Pearson’s correlations or goodness -of-fit